$F(t)=\int^\infty_0 {x^2}{e^{-tx}} dx$ is a continuous function?

Question

$F(t)=\int^\infty_0 {x^2}{e^{-tx}} dx$ defined for $t>0$ is a continuous function? How prove? I tried to use the definition of continuity, but I could not. Do you have another method?

Answer 1

You can either do integration by parts to find a better expression for the function or use the dominated convergence theorem, If you are familiar with measure theory.

Answer 2

$${\displaystyle\int}x^2\mathrm{e}^{-tx}\,\mathrm{d}x$$ Using integration by parts, $\int fg' = fg - \int f'g$ where $f = x^2$ and $g = e^{-tx}$, you can show that $${\displaystyle\int}x^2\mathrm{e}^{-tx}\,\mathrm{d}x =-\dfrac{x^2\mathrm{e}^{-tx}}{t}-{\displaystyle\int}-\dfrac{2x\mathrm{e}^{-tx}}{t}\,\mathrm{d}x $$ Notice that $${\displaystyle\int}-\dfrac{2x\mathrm{e}^{-tx}}{t}\,\mathrm{d}x =-{{\dfrac{2}{t}}}{\displaystyle\int}x\mathrm{e}^{-tx}\,\mathrm{d}x$$ Using integration by parts, $\int fg' = fg - \int f'g$ where $f = x$ and $g = e^{-tx}$, you can show that

\begin{equation} \int x\mathrm{e}^{-tx} =-\dfrac{x\mathrm{e}^{-tx}}{t}+{\displaystyle\int}\dfrac{\mathrm{e}^{-tx}}{t}\,\mathrm{d}x = -\dfrac{x\mathrm{e}^{-tx}}{t}-\frac{1}{t^2}e^{-tx} \end{equation} Going back to the first equation, we get $${\displaystyle\int}x^2\mathrm{e}^{-tx}\,\mathrm{d}x = -\dfrac{\left(t^2x^2+2tx+2\right)\mathrm{e}^{-tx}}{t^3}+A$$ Including the limits, we get $$\int\limits_0^{\infty}x^2\mathrm{e}^{-tx}\,\mathrm{d}x = -\dfrac{\left(t^2x^2+2tx+2\right)\mathrm{e}^{-tx}}{t^3}\Big\vert_0^{\infty} = \frac{2}{t^3}$$ So, now you have to check if $\frac{2}{t^3}$ is continuous on $]0,\infty[$.

Answer 3

Substituting $u=tx$, we have $$ F(t) = \int_0^\infty x^2e^{-tx}\,dx = \frac{1}{t^2} \int_0^\infty (tx)^2e^{-tx} \,dx = \frac1{t^2} \int_0^\infty u^2e^{-u} \frac{du}t = \frac{F(1)}{t^3} $$ so if you can prove that $F(t)$ always exists, it will be as continuous as $t^{-3}$.

Answer 4

For fixed $t$, let $$Ｃ=\int^\infty_0 {x^2}e^{-tx}dx.$$ Choose $\Delta t\in R$ such that $t+\Delta t>0$ and $$ |e^{-x\Delta t}-1|\le \frac1C|\Delta t| $$ and hence \begin{eqnarray} |F(t+\Delta t)-F(t)|&=&\left|\int^\infty_0 {x^2}(e^{-(t+\Delta t)x}-e^{-t}) dx\right|\\ &=&\left|\int^\infty_0 {x^2}e^{-tx}(e^{-x\Delta t}-1) dx\right|\\ &\le&\int^\infty_0 {x^2}e^{-tx}\left|e^{-x\Delta t}-1\right|dx\\ &\le&C|\Delta t|\int^\infty_0 {x^2}e^{-tx}dx\\ &\le&|\Delta t| \end{eqnarray} which implies $$ \lim_{\Delta t\to0}F(t+\Delta t)=F(t).$$ So $F(t)$ is continuous.